Chem II Homework Page, Exam 4 Material

This page has all of the required homework for the material covered in the fourth exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Equilibrium (Ch. 15-17)

The gaseous reaction

2NO(g) + O2(g) ⇔ 2NO2

has Kc = 1.20 at 1000 K. At equilibrium it was found that there were 7.50 g NO and 26.67 g O2 in a 250 mL container. What is the concentration of NO2 at this equilibrium?

A mixture of 0.03 moles of PCl3 and 0.02 moles of Cl2 react in a 100 mL container as follows:

PCl3(g) + Cl2(g) ⇔ PCl5(g)

At equilibrium there are 0.019 moles of PCl3 present in the container. (a) Calculate the equilibrium concentrations of PCl3, Cl2, and PCl5. (b) Calculate Kc under these conditions.

Answer

PCl3(g) + Cl2(g) ⇔ PCl5(g)

Init. Moles

0.03 mole

0.02 mole

0 mole

Init. Conc.

0.3 M

0.2 M

0 M

Equilib

0.3 - x

0.2 - x

x

Equilib

0.19

?

?

(a) Find x: 0.3 - x = 0.19; x = 0.3 - 0.19 = 0.11

⇒ [PCl5] = x = 0.11 M

⇒ [Cl2] = 0.2 - x = 0.2 - 0.11 = 0.09 M

[PCl3] = 0.19 M (given in the problem)

(b) Kc = [PCl5]/{[PCl3][Cl2]} = (0.11 M)/{(0.19 M)(0.09 M)} = 6.43

Nitrogen gas, N2(g), and water vapor react as follows:

N2(g) + 2H2O(g) ⇔ 2NO(g) + 2H2(g)

When 0.2 moles of N2 are combined with 0.3 moles of H2 in a 1.0 L container and allowed to go to equilibrium, it is found that there are 0.169 moles of N2 in the equilibrium mixture. What are the equilibrium concentrations of each of the materials and what is the value of the equilibrium constant, Kc?

Answer

N2(g) + H2O(g) ⇔ 2NO(g) + 2H2(g)

Init. Moles

0.2 mole

0.3 mole

0 mole

0 mole

Init. Conc.

0.2 M

0.3 M

0 M

0 M

Equilib

0.2 - x

0.3 - 2x

2x

2x

Equilib

0.169

?

?

?

(a) Find x: 0.2 - x = 0.169; x = 0.2 - 0.169 = 0.031

⇒ [H2O] = 0.3 - 2x = 0.238 M

⇒ [H2] = [NO] = 2x = (2)(0.031) = 0.062 M

[N2] = 0.169 M (given in the problem)

(b) Kc = [NO]²[H2]²/{[N2][H2O]²}

Kc = (0.062 M)²(0.062 M)²/{(0.169 M)(0.238 M)²} = 1.54 x 10-3

Consider the following reaction:

2NO(g) + Cl2(g) ⇔ 2NOCl(g) Kc = 1.42 x 107

Calculate the concentration of NO at equilibrium when 3.0 moles of NO are reacted with 2.0 moles of Cl2 in a 500 mL container.

Answer

A large Kc means that the reaction will go to right in the complete reaction line below.

2NO(g) + Cl2(g) ⇔ 2NOCl(g)

Init. Moles

3.0 mole

2.0 mole

0 moles

Init. Conc.

6 M

4 M

0 M

Complete Rxn

0

1

6

Equilib Conc

0 + 2x

1 + x

6 - 2x

Kc = 1.42 x 107 = [NOCl]²/{[NO]²[Cl2]} = (6-2x)²/{(2x)²(1+x)}

Assume x << 1

⇒ x = {36/[(4)(1.42 x 107)]}½ = 8.02 x 10−4

⇒ [NO] = 2x = 16 x 10−4 M

Consider the following reaction:

2NOCl(g) ⇔ 2NO(g) + Cl2(g) Kc = 7.04 x 10-8

Calculate the concentration of NO at equilibrium when 3.0 moles of NOCl and with 2.0 moles of Cl2 are initially placed in a 500 mL container.

Answer

A small Kc means that the reaction will go to left in the complete reaction line below.

2NOCl(g) ⇔ 2NO(g) + Cl2(g)

Init. Moles

3.0 mole

0 mole

2.0 moles

Init. Conc.

6 M

0 M

4 M

Complete Rxn

6

0

4

Equilib Conc

6 - 2x

0 + 2x

4 + x

Kc = 7.04 x 10−8 = {[NO]²[Cl2]}/[NOCl]² = {(2x)²(4+x)}/(6-2x)²

Assume x << 4

⇒ x = {(36)(7.04 x 10−8}/[(4)(4)]}½ = (1.584 x 10−7}½ = 3.98 x 10−4

⇒ [NO] = 2x = 7.96 x 10−4 M

Calculate the pH when 132 g of butanoic acid, HC4H7O2, which has Ka = 1.5 x 10-5, is placed in 300 mL of water.

Answer

Ka = 1.5 x 10-5 means that the reaction will go toward the left in the complete reaction line below.

HC4H7O2⇔ H+ + C4H7O2-

Init. Moles

1.5 mole

0 mole

0 mole

Init. Conc.

5 M

0 M

0 M

Complete Rxn

5

0

0

Equilib Conc

5 - x

x

x

Ka = 1.5 x 10-5 = [H+][C4H7O2-]/[HC4H7O2] = x²/(5 - x) Assume x << 5

⇒ x = [(5)(1.5 x 10-5)]½ = 0.00866 = [H+]

⇒ pH = -log(0.00866) = 2.06

Calculate the pH in a solution where 1.2 moles of formic acid, HCHO2, and 2.4 moles of sodium formate, NaCHO2, are mixed with water to make a total of 2.0 L of solution. For formic acid, Ka = 1.8 x 10-4.

Answer

Ka = 1.8 x 10-4 means that the reaction will go toward the left in the complete reaction line below. Notice that the Na+ ion is a spectator ion and doesn't show up in the equilibrium equation.

HCHO2⇔ H+ + CHO2-

Init. Moles

1.2 mole

0 mole

2.4 mole

Init. Conc.

0.6 M

0 M

1.2 M

Complete Rxn

0.6

0

1.2

Equilib Conc

0.6 - x

x

1.2 + x

Ka = 1.8 x 10-4 = [H+][CHO2-]/[HCHO2] = (x)(1.2 + x)/(0.6 - x)

Assume x << 0.6

⇒ x = [(0.6)(1.8 x 10-4)]/1.2 = 9 x 10-5 = [H+]

⇒ pH = -log(9 x 10-5) = 4.05

Determine the concentration of all ions in a solution where solid calcium hydroxide, Ca(OH)2, is in equilibrium with its ions. For calcium hydroxide Ksp = 6.5 x 10-6.

Answer

Ksp = 6.5 x 10−6 means that there will not be very many ions in the solution. Because the solid (concentration set to one) is in equilibrium with the ions, we only need to consider what the condition is at equilibrium as shown in the table.

Ca(OH)2(s) ⇔ Ca2+ + 2OH-

Equilib Conc

-

x

2x

Ksp = 6.5 x 10-6 = [Ca2+][OH−]² = (x)(2x)² = 4x³

⇒ x = [(6.5 x 10-6)/4]&frac13; = 0.0118

⇒[Ca2+] = 0.0118 M and [OH-] = 0.0236 M

An 8.5 mL sample of 1.0 M NaOH was needed to reach the equivalence point when titrating into 6.0 mL of acetic acid, HC2H3O2. Acetic acid has Ka = 1.8 x 10−5. What is the pH of the original acetic acid solution?

Answer

(0.0085 L NaOH)(1.0 mole/L NaOH) = 0.0085 moles of OH−

NaOH reacts with acetic acid in a one-to-one ratio. The NaOH reacts with both the dissociated and nondissociated hydrogen ions from the acetic acid solution, so the moles of NaOH used is the initial moles of HC2H3O2 that would have been in the solution before any dissociation of the acid.

Ka = 1.8 x 10-5 means that the reaction will go toward the left in the complete reaction line below.